faktorisieren p^6q^9+r^3

Lösung

faktorisieren

p^{6} q^{9} + r^{3}

(p^{2} q^{3} + r) (p^{4} q^{6} - p^{2} q^{3} r + r^{2})

Schritte zur Lösung

p^{6} q^{9} + r^{3}

Schreibe

p^{6} q^{9}

um:

(p^{2} q^{3})^{3}

= (p^{2} q^{3})^{3} + r^{3}

Wende Formel zur Summe von dritten Potenzen an:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(p^{2} q^{3})^{3} + r^{3} = (p^{2} q^{3} + r) ((p^{2} q^{3})^{2} - p^{2} q^{3} r + r^{2})

= (r + p^{2} q^{3}) ((p^{2} q^{3})^{2} - r p^{2} q^{3} + r^{2})

Vereinfache

(p^{2} q^{3} + r) ((p^{2} q^{3})^{2} - p^{2} q^{3} r + r^{2}) : (p^{2} q^{3} + r) (p^{4} q^{6} - p^{2} q^{3} r + r^{2})

= (p^{2} q^{3} + r) (p^{4} q^{6} - p^{2} q^{3} r + r^{2})

f a c t or 4 r^{2} - π r^{2}

f a c t or 3 (x + y)^{2} - 4 (x + y) - 4

f a c t or 5 (x + y)^{2} - 17 (x + y) + 6

s im pl i f y \frac{63}{1 + 8 e ^{- 315 (5)}}

\frac{d}{d t} ((x) (t^{2}))